A Singular Meixner-Process Integral That Diverges

Summary

The document asks how to evaluate an integral involving an exponential term divided by the product of the variable and a hyperbolic sine, in the context of numerical analysis for the Meixner Lévy process. The response shows that the integral over the full real line does not converge, so numerical integration cannot produce a meaningful finite value as posed.

Near zero, the hyperbolic sine has a linear leading term. After multiplication by the variable, the denominator behaves quadratically, while the exponential numerator approaches a finite nonzero value. The integrand therefore has a non-integrable singularity proportional to the inverse square of the variable. The answer further states that taking the Cauchy principal value does not resolve this divergence. It does not propose a regularization, modified integrand, or alternative numerical formulation.

Key ideas

• The stated integral over the real line diverges because of its behavior near zero.
• Near the singularity, the integrand behaves like a constant times the inverse square of the variable.
• The singularity is not removed by taking the Cauchy principal value.
• The response does not offer a regularized alternative for Meixner-process calculations.

Tags

# Integration in the context of modelling with the Meixner Process


# Integration in the context of modelling with the Meixner Process












I failed to evaluate the integral of $\frac{e^{ax}}{x\sinh(bx)}$ with respect to $x$ from negative infinite to positive infinite. What techniques can I use to evaluate the integrals of such kind for the Meixner Levy process for the purpose of numerical analysis?

## Answer by olaker (score 2)

https://quant.stackexchange.com/a/10865

The integral diverges, hence numerical integration will not yield a meaningful result.

Indeed, using the Taylor expansion of the hyperbolic sine one has that $$x \sinh (bx)= x\frac{e^{bx}-e^{-bx}}{2}=bx^2+\frac{b^3x^4}{6}+...$$ Therefore, at the vicinity of $x=0$ $$\frac{e^{ax}}{x \sinh (bx)}\sim\frac{1}{bx^2}$$ which is a non-integrable singularity even in terms of the Cauchy principal value.

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